Notes: Exponential Equations (4.5–4.6)

Exponential equations: core ideas

  • Base equality rule: If bases are the same and positive (a > 0, a ≠ 1), then
    af(x)=ag(x)⇒f(x)=g(x).a^{f(x)} = a^{g(x)} \Rightarrow f(x) = g(x).
  • When bases are different or the exponent contains the variable, use logarithms to isolate the exponent.
  • If you have an expression like A=B ektA = B \, e^{kt}, isolate the exponential: AB=ekt⇒t=ln⁡(A/B)k.\frac{A}{B} = e^{kt} \Rightarrow t = \frac{\ln(A/B)}{k}.
  • For equations with base e specifically, use natural log (ln); for base 10 use common log (log). The same idea applies with any log: log⁡a(B)=ln⁡(B)ln⁡(a).\log_a(B) = \frac{\ln(B)}{\ln(a)}.

Quick ruleset for solving exponential equations

  • If the equation can be written as a single base: use the base-equality rule to equate exponents.
  • If not, take logs of both sides and solve for the unknown.
  • Always check simple linear-exponent steps by back-substitution when possible.

Worked quick problems (4.5)

  • Problem: 3=3w\sqrt{3} = 3^w
    • Rewrite left as a power of 3: 3=31/2\sqrt{3} = 3^{1/2}
    • Therefore w=12.w = \frac{1}{2}.
  • Problem: 21000=63000 e−0.2t21000 = 63000 \, e^{-0.2 t}
    • Divide by 63000: 13=e−0.2t\frac{1}{3} = e^{-0.2 t}
    • Take natural log: ln⁡(13)=−0.2t\ln\left(\frac{1}{3}\right) = -0.2 t
    • Solve for t: t=ln⁡(1/3)−0.2=ln⁡30.2=5ln⁡3.t = \frac{\ln(1/3)}{-0.2} = \frac{\ln 3}{0.2} = 5 \ln 3. (Optional: numerical value ≈ 5.49)

Additional notes on problem-solving flow

  • When the base is the same on both sides, aim to equate exponents first to get a linear equation in the unknown.
  • When the exponent is in terms of a variable and the bases are not easily matched, apply a logarithm to both sides and solve for the variable.
  • For equations of the form A=BektA = B e^{kt}, use the steps above to isolate t via a natural log.

Quick reference formulas

  • If au=ava^{u} = a^{v} with a>0,a≠1a>0, a\neq 1, then u=v.u = v.
  • If A=BuA = B^u, then u=log⁡BA=ln⁡Aln⁡B.u = \log_B A = \frac{\ln A}{\ln B}.
  • If A=BektA = B e^{kt}, then t=ln⁡(A/B)k.t = \frac{\ln(A/B)}{k}.